Surface-charge of a uniformly polarized sphere

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SUMMARY

The discussion focuses on the calculation of the total bound surface charge on the inner surface of a uniformly polarized sphere. The electric field inside the cavity is determined to be E = (-1/3*e_o)*P, where P represents polarization. The bound surface charge density is given by P*cos(theta), and the total charge must be calculated through integration rather than simple multiplication. The conclusion is that the total charge on the inner surface is zero due to the symmetry of the bound charges in the northern and southern hemispheres.

PREREQUISITES
  • Understanding of electric fields and polarization in dielectric materials.
  • Familiarity with spherical coordinates and integration techniques.
  • Knowledge of bound surface charge concepts in electrostatics.
  • Proficiency in calculus, particularly double integrals.
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  • Study the derivation of electric fields in polarized materials using Gauss's Law.
  • Learn about the implications of bound charges in electrostatics.
  • Explore advanced integration techniques in spherical coordinates.
  • Investigate the properties of dielectric materials under external electric fields.
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This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and researchers focusing on electrostatics and material properties of dielectrics.

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Homework Statement


Hi all.

Please take a look at #2 in http://www.physics.utoronto.ca/~hori/Courses/P352F/hw6.pdf

(this is not homework, I'm just training).

Ok, I can find the electric field inside the cavity y superposition and it is E = (-1/ 3*e_o)*P. I know that the bound surface charge is P*cos(theta), where theta is the usual spherical coordinate. My question is: Is the total charge on the inner surface P*cos(theta)*4*Pi*(R_0)^2?
 
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Where is [tex]\theta[/tex] in your last sentence? In order to find the total charge, you must integrate - not multiply - since the surface charge varies with respect to [tex]\theta[/tex]. The proper way to find the surface charge would be

[tex]Q = \int_S \sigma(\theta) \, da = \int_0^{2\pi} d\phi \int_0^{\pi} \cos \theta \sin \theta d\theta = 0[/tex]

It's zero, because on the northern hemisphere the bound charge is positive, which is compensated by the symmetrically negative bound charge on the southern hemisphere.

In conclusion, the answer is no.
 

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